Math Help - Another Question regarding Sequences and Series

1. Another Question regarding Sequences and Series

Hello everyone. I'm very thankful for all who have helped me before, and sorry about asking another question in that thread, as well as bombarding you guys with questions. Anways, i got some really weird answer for this question, and I'm hoping that someone can show me how to do it. Thank you all!

Find n for a sequence for which a=-7, d=1.5, and Sn=-14.

2. Let us see.

a=-7, d=1.5, and Sn=-14.
I assume a=-7 is actually a1 = -7

So,
an = a1 +(n-1)d
an = -7 +(n-1)(1.5)
an = 1.5n -8.5 -----------(i)

And,
Sn = (n/2)(a1 +an)
-14 = (n/2)(-7 +1.5n -8.5)
-28 = n(1.5n -15.5)
-28 = 1.5n^2 -15.5n
1.5n^2 -15.5n +28 = 0
n = {15.5 +,-sqrt[(15.5)^2 -4(1.5)(28)]} / 2(1.5)
n = {15.5 +,-8.5} /3
n = 8 or 2.333

Reject the n=2.333 because n is the number of terms in the sequence/series. So n has to be a positive integer always.

3. Hello, Norman!

Find $n$ for a sequence for which: . $a=-7,\:d=1.5,\:S_n=-14$
The sum of the first n terms of an arithmetic series is given by:

. . $S_n \;=\;\frac{n}{2}\left[2a + (n-1)d\right]$

Substitute the given values: . $-14 \;=\;\frac{n}{2}\left[2(-7) + (n-1)(1.5)\right]$

Multiply by 2: . $-28 \;=\;n\left(-14 + 1.5n - 1.5\right)\quad\Rightarrow\quad 1.5n^2 - 15.5n + 28 \;=\;0$

Multiply by 2: . $3n^2 - 31n + 56\;=\;0\quad\Rightarrow\quad(n-8)(3n-7) \:=\:0$

. . which has roots: . $n\:=\:8,\:\frac{7}{3}$

Since $n$ must be a positive integer: . $\boxed{n \,=\,8}$